Question: Let $f$ be a vector-valued function defined by $f(t)=(6t^2-1,-4\ln(t))$. Find $f$ 's second derivative $f''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(12t,-\dfrac4t\right)$ (Choice B) B $\left(6,-\dfrac4{t^2}\right)$ (Choice C) C $\left(12,\dfrac{4}{t^2}\right)$ (Choice D) D $6-\dfrac4{t^2}$
Solution: We are asked to find the second derivative of $f$. This means we need to differentiate $f$ twice. In other words, we differentiate $f$ once to find $f'$, and then differentiate $f'$ (which is a vector-valued function as well) to find $f''$. Recall that $f(t)=(6t^2-1,-4\ln(t))$. Therefore, $f'(t)=\left(12t,-\dfrac4t\right)$. Now let's differentiate $f'(t)=\left(12t,-\dfrac4t\right)$ to find $f''$. $f''(t)=\left(12,\dfrac{4}{t^2}\right)$ In conclusion, $f''(t)=\left(12,\dfrac{4}{t^2}\right)$.